Question

Figure gives the electric potential V(x) along a copper wire carrying uniform current, from a point of higher potential ${\mathbf{V}}_{\mathbf{s}}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{\mu V}$at x = 0to a point of zero potential at ${\mathbf{x}}_{\mathbf{s}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{m}$. The wire has a radius of . What is the current in the wire?

Short answer

The current in the wire is 2.97mA .

Step-by-step solution

Identification of given data

1. A higher potential at x = 0 is$V=12\mu V\mathrm{or}12\times {10}^{-6}\mathrm{V}$ .
2. A lower potential at${\mathrm{x}}_{\mathrm{s}}=3.00\mathrm{m}$ is V = 0 .
3. The radius of the wire, r = 2mm or 0.002m

Significance of Ohm’s law

Ohm’s law states that the current flowing between the two terminals is directly proportional to the potential difference between these two terminals. The opposition to the flow of current is called resistance.

By using Ohm’s law and the formula for resistance and substituting all the given values, we can find the current in the wire.

Formulae:

The voltage equation from Ohm’s law, V = IR …(i)

Here,

V is the potential difference between the two terminals, I is current, and R is the resistance between the two terminals.

The resistance of the material, $R=\frac{Lp}{A}$ …(ii)

Here, is the resistance between the two terminals, L is the length of the current, A is the area of cross-section of the conductor, p is the resistivity,

The cross-sectional area of the wire, $A=\pi r^2$ …(iii)

Here, is the area of cross-section of the conductor, r is the radius.

Determining the current in the wire

Substituting the value of area from equation (iii) in equation (ii), we can get the value of the resistance of the wire as follows:

$R=\frac{Lp}{\pi r^2}$

Now, substituting the above value of the resistance in equation (i), we can get the value of the current in the wire by using the given data as follows:

(The value of resistivity for copper is$\mathrm{p}=1.69\times {10}^{-8}\mathrm{\Omega }.\mathrm{m}$ )

$$\begin{gathered} \mathrm{I}=\frac{{\mathrm{V\pi r}}^2}{\mathrm{Lp}} \\ =\frac{\left(12\times {10}^{-6}\mathrm{V}\right)\times \mathrm{\pi }\times {\left(0.002\mathrm{m}\right)}^2}{\left(3.00\mathrm{m}\right)\times \left(1.69\times {10}^{-8}\mathrm{\Omega }.\mathrm{m}\right)} \\ =2.97\times {10}^{-3}\mathrm{A} \\ =2.97\mathrm{mA} \end{gathered}$$

Hence, the value of the current is 2.97mA .